142.linked-list-cycle-ii.go

/*
 * @lc app=leetcode id=142 lang=golang
 *
 * [142] Linked List Cycle II
 *
 * https://leetcode.com/problems/linked-list-cycle-ii/description/
 *
 * algorithms
 * Medium (39.63%)
 * Likes:    3794
 * Dislikes: 287
 * Total Accepted:    419.3K
 * Total Submissions: 1.1M
 * Testcase Example:  '[3,2,0,-4]\n1'
 *
 * Given a linked list, return the node where the cycle begins. If there is no
 * cycle, return null.
 *
 * There is a cycle in a linked list if there is some node in the list that can
 * be reached again by continuously following the next pointer. Internally, pos
 * is used to denote the index of the node that tail's next pointer is
 * connected to. Note that pos is not passed as a parameter.
 *
 * Notice that you should not modify the linked list.
 *
 *
 * Example 1:
 *
 *
 * Input: head = [3,2,0,-4], pos = 1
 * Output: tail connects to node index 1
 * Explanation: There is a cycle in the linked list, where tail connects to the
 * second node.
 *
 *
 * Example 2:
 *
 *
 * Input: head = [1,2], pos = 0
 * Output: tail connects to node index 0
 * Explanation: There is a cycle in the linked list, where tail connects to the
 * first node.
 *
 *
 * Example 3:
 *
 *
 * Input: head = [1], pos = -1
 * Output: no cycle
 * Explanation: There is no cycle in the linked list.
 *
 *
 *
 * Constraints:
 *
 *
 * The number of the nodes in the list is in the range [0, 10^4].
 * -10^5 <= Node.val <= 10^5
 * pos is -1 or a valid index in the linked-list.
 *
 *
 *
 * Follow up: Can you solve it using O(1) (i.e. constant) memory?
 *
 */
package leetgo

// @lc code=start
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
	if head == nil {
		return nil
	}
	slow, fast := head, head

	for fast != nil && fast.Next != nil {
		fast = fast.Next.Next
		slow = slow.Next

		if fast == slow {
			break
		}
	}
	if fast == nil || fast.Next == nil {
		return nil
	}

	slow = head

	for slow != fast {
		slow = slow.Next
		fast = fast.Next
	}

	return fast
}

// @lc code=end